# Isomorphic representations on exterior powers

Exercise from F+H, Exercise 1.3:

Let $\rho : G \rightarrow GL(V)$ be any representation of the finite group $G$ on a $n$-dimensional vector space $V$ and suppose that for any $g \in G$ the determinant of $\rho(g)$ is 1. Show that $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$ are isomorphic as representations of G

For the life of me I can't figure out. I know that $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$ are isomorphic as spaces, but why are they isomorphic as representations, I have no idea. I suspect it has something to do with the determinant being 1, but...

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If you know they are isomorphic as spaces, can you write down an isomorphism between them? Can you do it without choosing a basis? If so, your isomorphism is probably an isomorphism of representations. Hint: show that they are both the dual of $\Lambda^k V^{\ast}$ (as representations). – Qiaochu Yuan Nov 14 '12 at 20:00

For any field $F$ and vector space $V$ over $F$ there is a map $$\wedge^k V \rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, \wedge^n V)$$ defined by sending an element $x \in \wedge^k V$ to the linear map $y \mapsto x \wedge y$ which is wedge product with $x$. If $V$ is $n$-dimensional then the determinant induces an isomorphism of vector spaces $$\wedge^n V \rightarrow F$$ so that by composing we obtain a map of vector spaces $$\phi:\wedge^k V \rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, \wedge^n V)\rightarrow \mathrm{Hom}_F(\wedge^{n-k} V, F)$$ that you can show is actually an isomorphism. For any $g \in GL(V)$ we have $\phi(gx)=\mathrm{det}(g) g \phi(x)$, so if $\mathrm{det}(g)=1$ this reads $\phi(gx)=g\phi(x)$, as desired.